Answer:
∴ Q = -7.52kCal
Explanation:
Using the formula for specific heat capacity:
Q = mcΔT
where ΔT = change in temperature (final - initial) = (0 - 100)°C = -100°C
m = mass (g) = 75g
c = specific heat capacity = 4.2 J/g°C in water
⇒ Q = 75 × 4.2 × -100
= -31,500J
But 1J - 0.000239kCal
<u>∴ Q = -7.52kCal</u>
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Let me know if I can be of further assistance.
Blue has a shorter wave length
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
Theyre temporary magnets, so they can be turned on and off.
